Question

if kinetic energy of a particle is decreased by 19% by what percent is its Momentum is change

Answer 1

K.E=1/2mv^2
P=mv
OR
v=P/m(rearranged)
K.E=1/2m(P/m)^2
K.E=1/2m (P^2/m^2)

this becomes K.E=P^2/2m
If mass is constant then we can say

K.E is directly proportional toP^2
Consider a body having kinetic energy K1 and P1.lets it kinetic energy change to K2 and momentum to P2.
Then
K1/K2=(P1^2/P2^2)
since K2 is 19% less than K1
then K2=K1-19%K1
K1/K1-19%K1=(P1/P2)^2
K1/K1-19/100K1=(P1/P2)^2
K1/K1-0.19K1=(P1/P2)^2
K1/0.81K1=(P1/P2)^2
K1 and K1 cancel out in numerator and denominator
1/0.81=(P1/P2)^2
taking square root on b/s
10/9=P1/P2
10P2=9P1
P2=9/10 P1
P2=0.9P1

P2-P1=0.9P1-P1
          =-0.1P1
calculating fractional change
P2-P1/P1=-0.1
calculating the percentage change in momentum
%P2_P1/P1=-0.1×100
                     =-10%
hope it helps u
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